Question: If $f'(x)=1/f(x)$ and $f(0)=2$, then $f(4)$ can be written in fully simplified form as $m\sqrt n$ for some integers $m$ and $n$. What are $m$ and $n$ ? $m=~$
Explanation: If we let $y=f(x)$ and change the derivative notation, we can rewrite the differential equation so that it's separable. $\dfrac{dy}{dx}=\dfrac1y$ What does it look like after we separate the variables? $y\,dy=dx$ Let's integrate both sides of the equation. $\int y\,dy=\int dx$ What do we get? $\dfrac{y^2}2=x+C$ What value of $C$ satisfies the initial condition $y(0)=2$ ? Let's substitute $x=0$ and $y=2$ into the equation and solve for $C$. $\begin{aligned} \dfrac{2^2}2&=0+C\\ \\ \\ C&=2 \end{aligned}$ Now use this value of $C$ to express $y$ in terms of $x$. $\begin{aligned} \dfrac{y^2}2&=x+2\\ \\ \\ y^2&=2x+4\\ \\ y&=\pm\sqrt{2x+4} \end{aligned}$ Which sign satisfies the initial condition $y(0)=2$ ? We must choose the positive sign to satisfy the initial condition. $ y=\sqrt{2x+4}$ If $y(4)=m\sqrt n$ in fully simplified form, what are the integers $m$ and $n$ ? $\begin{aligned} y(4)&=\sqrt{2\cdot4+4}\\ \\ &=\sqrt{3\cdot4}\\ \\ &=2\sqrt3\\ \\ &=m\sqrt n \end{aligned}$ Thus, $m=2$ and $n=3$.